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Sample Mock Test JEE
15 questions · Physics, Chemistry, Maths · tap an option to check your answer
Physics
Q1. The efficiency of a Carnot engine operating between temperatures of 100°C and −23°C will be?
Answer: B
The efficiency of a Carnot engine is given by \( \eta = 1 - \frac{T_2}{T_1} \). Here, \( T_1 = 100 + 273 = 373 \) K and \( T_2 = -23 + 273 = 250 \) K. Substituting these values, we find \( \eta = 1 - \frac{250}{373} = \frac{123}{373} \). Thus, the efficiency is \( \frac{100 + 23}{373} \).
The efficiency of a Carnot engine is given by \( \eta = 1 - \frac{T_2}{T_1} \). Here, \( T_1 = 100 + 273 = 373 \) K and \( T_2 = -23 + 273 = 250 \) K. Substituting these values, we find \( \eta = 1 - \frac{250}{373} = \frac{123}{373} \). Thus, the efficiency is \( \frac{100 + 23}{373} \).
Q2. The intensity at the maximum in a Young’s double slit experiment is \( I_0 \). Distance between two slits is \( d=5\lambda \), where \( \lambda \) is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance \( D= 10d \)?
Answer: C
Given that the distance between slits \( d = 5\lambda \) and the distance from the slits to the screen is \( D=10d = 50\lambda \). The intensity at the maximum in this Young’s double slit experiment is \( I_0 \). The path difference \( \Delta y = \frac{dY}{D} = \frac{5\lambda Y}{50\lambda} = \frac{Y}{10} \). For a path difference of \( \frac{\lambda}{4} \), we find the phase difference \( \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \). Thus, the intensity is given by \( I = I_0 \cos^2 \left( \frac{\phi}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{4} \right) = \frac{I_0}{2} \). Therefore, the intensity in front of one of the slits is \( \frac{I_0}{2} \).
Given that the distance between slits \( d = 5\lambda \) and the distance from the slits to the screen is \( D=10d = 50\lambda \). The intensity at the maximum in this Young’s double slit experiment is \( I_0 \). The path difference \( \Delta y = \frac{dY}{D} = \frac{5\lambda Y}{50\lambda} = \frac{Y}{10} \). For a path difference of \( \frac{\lambda}{4} \), we find the phase difference \( \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \). Thus, the intensity is given by \( I = I_0 \cos^2 \left( \frac{\phi}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{4} \right) = \frac{I_0}{2} \). Therefore, the intensity in front of one of the slits is \( \frac{I_0}{2} \).
Q3. The escape velocity from the earth is 11.2 km/s. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is?
Answer: B
Escape velocity does not depend on the angle of projection. It remains the same, which is 11.2 km/s.
Escape velocity does not depend on the angle of projection. It remains the same, which is 11.2 km/s.
Q4. The force on a rocket moving with a velocity of 300 m/s is 210 N. The rate of consumption of fuel of the rocket is?
Answer: A
Thrust force on the rocket is given by the relation \( F = -v \frac{dm}{dt} \). Given \( F = 210 \text{ N} \) and \( v = 300 \text{ m/s} \), we can solve for \( \frac{dm}{dt} \):\n\[ -\frac{dm}{dt} = \frac{F}{v} = \frac{210}{300} = 0.7 \text{ kg/s} \]\nThus, the rate of combustion of fuel is 0.7 kg/s.
Thrust force on the rocket is given by the relation \( F = -v \frac{dm}{dt} \). Given \( F = 210 \text{ N} \) and \( v = 300 \text{ m/s} \), we can solve for \( \frac{dm}{dt} \):\n\[ -\frac{dm}{dt} = \frac{F}{v} = \frac{210}{300} = 0.7 \text{ kg/s} \]\nThus, the rate of combustion of fuel is 0.7 kg/s.
Q5. A particle of mass \( m \) is driven by a machine that delivers a constant power \( k \) watts. If the particle starts from rest, the force on the particle at time \( t \) is?
Answer: A
As the machine delivers a constant power, \( P = Fv \). Thus, \( F = \frac{P}{v} = k/v \). Integrating gives us the velocity as a function of time, leading to \( v = \frac{k}{m}t^{1/2} \). The force on the particle is then given by \( F = m \frac{dv}{dt} = mk t^{-1/2} \).
As the machine delivers a constant power, \( P = Fv \). Thus, \( F = \frac{P}{v} = k/v \). Integrating gives us the velocity as a function of time, leading to \( v = \frac{k}{m}t^{1/2} \). The force on the particle is then given by \( F = m \frac{dv}{dt} = mk t^{-1/2} \).
Chemistry
Q6. Bleaching powder reacts with a few drops of conc. HCl to give
Answer: A
When bleaching powder (Ca(OCl)₂) reacts with hydrochloric acid (HCl), it produces chlorine gas (Cl₂). The reaction is: Ca(OCl)₂ + 2 HCl → CaCl₂ + H₂O + Cl₂.
When bleaching powder (Ca(OCl)₂) reacts with hydrochloric acid (HCl), it produces chlorine gas (Cl₂). The reaction is: Ca(OCl)₂ + 2 HCl → CaCl₂ + H₂O + Cl₂.
Q7. The central dogma of molecular genetics states that the genetic information flows from
Answer: C
The central dogma of molecular genetics states that DNA is transcribed into RNA, which is then translated into proteins. Thus, the correct flow is DNA → RNA → proteins.
The central dogma of molecular genetics states that DNA is transcribed into RNA, which is then translated into proteins. Thus, the correct flow is DNA → RNA → proteins.
Q8. What is the chemical formula for water?
Answer: A
Water is composed of two hydrogen atoms bonded to one oxygen atom, giving it the chemical formula H2O. This can be derived from the molecular composition of water, where the ratio of hydrogen to oxygen is 2:1.
Water is composed of two hydrogen atoms bonded to one oxygen atom, giving it the chemical formula H2O. This can be derived from the molecular composition of water, where the ratio of hydrogen to oxygen is 2:1.
Q9. The slope of Arrhenius plot of first order reaction is
Answer: A
The slope of the Arrhenius plot (ln k vs 1/T) is given by the formula: \( -\frac{E_a}{R} \). Here, \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \), so we can calculate the activation energy \( E_a \) using the slope provided. The slope is given as \( -\frac{510}{3} \text{ K} \). Rearranging gives \( E_a = -\text{slope} \times R \). Calculating gives \( E_a = 41.5 \text{ kJ mol}^{-1} \). Thus, the correct option is (a) 41.5 kJ mol−1.
The slope of the Arrhenius plot (ln k vs 1/T) is given by the formula: \( -\frac{E_a}{R} \). Here, \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \), so we can calculate the activation energy \( E_a \) using the slope provided. The slope is given as \( -\frac{510}{3} \text{ K} \). Rearranging gives \( E_a = -\text{slope} \times R \). Calculating gives \( E_a = 41.5 \text{ kJ mol}^{-1} \). Thus, the correct option is (a) 41.5 kJ mol−1.
Q10. Number of chiral carbon atoms in β-D-(+)-glucose is
Answer: A
The number of chiral carbon atoms in β-D-(+) glucose are five.
The number of chiral carbon atoms in β-D-(+) glucose are five.
Maths
Q11. The number of integral values of \( n \) so that \( \sin(x) + \sin(\cos(x)) = n \) has at least one solution is
Answer: C
We have \( \sin(x) + \sin(\cos(x)) = n \) which leads to \( -2 \leq n \leq 2 \). Thus, there are two integral values of \( n \), which are 0 and 1.
We have \( \sin(x) + \sin(\cos(x)) = n \) which leads to \( -2 \leq n \leq 2 \). Thus, there are two integral values of \( n \), which are 0 and 1.
Q12. If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
Answer: B
To find the rank of the word KRISNA, we first arrange the letters in alphabetical order: A, I, K, N, R, R, S. We then calculate how many words would precede the arrangement of KRISNA by fixing each letter from the left and counting the permutations of remaining letters. The rank of KRISNA is found to be 341 after considering all preceding combinations.
To find the rank of the word KRISNA, we first arrange the letters in alphabetical order: A, I, K, N, R, R, S. We then calculate how many words would precede the arrangement of KRISNA by fixing each letter from the left and counting the permutations of remaining letters. The rank of KRISNA is found to be 341 after considering all preceding combinations.
Q13. The range of the function \( f(x) = \frac{x + 1}{x + 2} \) is
Answer: D
The function \( f(x) = \frac{x + 1}{x + 2} \) approaches 1 as \( x \to \infty \) but never actually reaches it. Therefore, the range is all real numbers except 1, which matches option D: None of these.
The function \( f(x) = \frac{x + 1}{x + 2} \) approaches 1 as \( x \to \infty \) but never actually reaches it. Therefore, the range is all real numbers except 1, which matches option D: None of these.
Q14. The locus of the vertices of the family of parabolas y=a(x^2-3a)+2 is
Answer: B
The given family of parabolas can be analyzed to find the locus of their vertices. By substituting various values of 'a' and simplifying, we find that the locus is given by xy=64/105.
The given family of parabolas can be analyzed to find the locus of their vertices. By substituting various values of 'a' and simplifying, we find that the locus is given by xy=64/105.
Q15. Find the mean deviation about median for the following data. Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of girls 6 8 14 16 42
Answer: A
To calculate the mean deviation about the median, first find the cumulative frequency and mid values for the given marks. The mid value for each class is calculated, and the absolute deviations from the median are then summed up and averaged. After performing the calculations, the correct mean deviation is found to be 10.34.
To calculate the mean deviation about the median, first find the cumulative frequency and mid values for the given marks. The mid value for each class is calculated, and the absolute deviations from the median are then summed up and averaged. After performing the calculations, the correct mean deviation is found to be 10.34.
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